Problem Description

You need to create a permutation of numbers from 1 to n such that the absolute differences between consecutive elements have exactly k distinct values.

Given two integers n and k, you must construct a list containing all integers from 1 to n (each appearing exactly once). The key requirement is that when you calculate the absolute differences between each pair of consecutive elements in your list, you should get exactly k distinct difference values.

For example, if your answer list is [a₁, a₂, a₃, ..., aₙ], then the differences list [|a₁ - a₂|, |a₂ - a₃|, |a₃ - a₄|, ..., |aₙ₋₁ - aₙ|] must contain exactly k unique values.

The solution uses a clever alternating pattern. For the first k elements, it alternates between picking from the smallest available number (starting from 1) and the largest available number (starting from n). This creates maximum differences initially. After placing k elements, the remaining elements are filled in sequential order (either ascending or descending based on whether k is even or odd) to ensure no new distinct differences are introduced.

The approach works because:

• Alternating between extremes (1, n, 2, n-1, 3, n-2...) creates differences of n-1, n-2, n-3, etc.
• After k alternations, we've created k distinct differences
• Filling the rest sequentially (all with difference 1) doesn't add new distinct values

Intuition

To get exactly k distinct differences, we need to think about how to control the variety of differences we create. The key insight is that we can maximize the number of distinct differences by alternating between the smallest and largest available numbers.

Consider what happens when we place numbers from opposite ends of the range:

• Placing 1 next to n gives us a difference of n-1
• Placing n next to 2 gives us a difference of n-2
• Placing 2 next to n-1 gives us a difference of n-3
• And so on...

This pattern naturally creates many distinct differences. If we alternate between picking from the left (1, 2, 3...) and right (n, n-1, n-2...) ends, we generate differences of n-1, n-2, n-3, etc. Each alternation creates a new unique difference value.

But we only want exactly k distinct differences, not more. So after making k alternations (which gives us k distinct differences), we need to stop creating new differences. How? By placing the remaining numbers consecutively - either all ascending or all descending. Consecutive numbers always have a difference of 1, which we've likely already created in our alternating phase.

The beauty of this approach is that after k alternations:

• If k is even, we last picked from the right side, so we continue with descending numbers
• If k is odd, we last picked from the left side, so we continue with ascending numbers

This ensures all remaining differences are 1, adding no new distinct values to our set of differences.

Learn more about Math patterns.

Solution Approach

The implementation uses a two-pointer technique with l starting at 1 and r starting at n. We build the answer array in two phases:

Phase 1: Create k distinct differences

For the first k positions (indices 0 to k-1), we alternate between picking from left and right:

• When index i is even: append l and increment l by 1
• When index i is odd: append r and decrement r by 1

This creates the pattern: [1, n, 2, n-1, 3, n-2, ...] which generates differences of n-1, n-2, n-3, and so on.

Phase 2: Fill remaining elements

For positions from k to n-1, we need to add the remaining numbers without creating new distinct differences. The strategy depends on the parity of k:

• If k is even: we last picked from the left pointer in Phase 1, so now we pick all remaining elements from the right pointer in descending order (r, r-1, r-2, ...)
• If k is odd: we last picked from the right pointer in Phase 1, so now we pick all remaining elements from the left pointer in ascending order (l, l+1, l+2, ...)

This ensures all consecutive differences in Phase 2 are 1, which is already one of our distinct differences from Phase 1.

Example walkthrough with n=5, k=3:

• Phase 1 (k=3 iterations):
  • i=0 (even): append 1, l becomes 2, ans = [1]
  • i=1 (odd): append 5, r becomes 4, ans = [1, 5]
  • i=2 (even): append 2, l becomes 3, ans = [1, 5, 2]
• Phase 2 (k is odd, so use left pointer):
  • append 3, l becomes 4, ans = [1, 5, 2, 3]
  • append 4, ans = [1, 5, 2, 3, 4]
• Differences: |1-5|=4, |5-2|=3, |2-3|=1, |3-4|=1
• Distinct differences: {4, 3, 1} - exactly 3 as required

Example Walkthrough

Let's walk through the solution with n = 6, k = 4.

We need to create a permutation of [1, 2, 3, 4, 5, 6] such that consecutive differences have exactly 4 distinct values.

Initial Setup:

• l = 1 (left pointer)
• r = 6 (right pointer)
• ans = [] (result array)

Phase 1: Create k=4 distinct differences

Iteration 1 (i=0, even):

• Pick from left: append 1
• Increment l to 2
• ans = [1]

Iteration 2 (i=1, odd):

• Pick from right: append 6
• Decrement r to 5
• ans = [1, 6]
• Difference created: |1-6| = 5

Iteration 3 (i=2, even):

• Pick from left: append 2
• Increment l to 3
• ans = [1, 6, 2]
• Difference created: |6-2| = 4

Iteration 4 (i=3, odd):

• Pick from right: append 5
• Decrement r to 4
• ans = [1, 6, 2, 5]
• Difference created: |2-5| = 3

Phase 2: Fill remaining elements without new differences

Since k=4 is even, we last picked from the left in Phase 1. Now we pick remaining elements from the right pointer in descending order:

Iteration 5:

• Append r = 4
• Decrement r to 3
• ans = [1, 6, 2, 5, 4]
• Difference created: |5-4| = 1

Iteration 6:

• Append r = 3
• ans = [1, 6, 2, 5, 4, 3]
• Difference created: |4-3| = 1

Final Result:

• Permutation: [1, 6, 2, 5, 4, 3]
• Consecutive differences: [5, 4, 3, 1, 1]
• Distinct differences: {5, 4, 3, 1} - exactly 4 distinct values ✓

The algorithm successfully creates exactly k=4 distinct differences by:

1. Alternating between extremes for the first 4 positions to create maximum variety
2. Filling the rest consecutively (descending) to avoid introducing new differences

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm consists of two sequential loops:

• The first loop runs k times, performing constant-time operations (appending to list and incrementing/decrementing pointers)
• The second loop runs n - k times, also performing constant-time operations

Since both loops together iterate exactly n times in total, and each iteration performs O(1) operations, the overall time complexity is O(k) + O(n - k) = O(n).

Space Complexity: O(n)

The space complexity is dominated by the output array ans which stores exactly n elements. The additional variables l, r, and i use constant space O(1). Therefore, the total space complexity is O(n) for the output array.

Note that if we consider only the auxiliary space (excluding the output array which is required for the answer), the space complexity would be O(1) as we only use a constant amount of extra space for the pointers and loop variables.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrectly Handling the Second Phase Direction

A frequent mistake is not properly determining which pointer (left or right) to use when filling the remaining elements after creating k distinct differences. The logic depends on which pointer was used last in Phase 1, not just the parity of k.

Incorrect Implementation:

# Wrong: Always using left pointer for remaining elements
for i in range(k, n):
    result.append(left)
    left += 1

Why it fails: If k is even, the last element in Phase 1 came from the left pointer (since we start with i=0 which is even). Continuing with the left pointer would create a difference of 1, but then switching directions later could introduce unwanted distinct differences.

Correct Implementation:

for i in range(k, n):
    if k % 2 == 0:
        # k even means last pick was from left, so continue from right
        result.append(right)
        right -= 1
    else:
        # k odd means last pick was from right, so continue from left
        result.append(left)
        left += 1

2. Off-by-One Error in Loop Boundaries

Another common mistake is using incorrect loop boundaries, particularly confusing whether to iterate k times or k+1 times for the first phase.

Incorrect Implementation:

# Wrong: Including index k in the alternating phase
for i in range(k + 1):  # This creates k+1 distinct differences!
    if i % 2 == 0:
        result.append(left)
        left += 1
    else:
        result.append(right)
        right -= 1

Why it fails: This creates k+1 positions with alternating pattern, potentially generating k+1 distinct differences instead of exactly k.

Correct Implementation:

# Iterate exactly k times to create k distinct differences
for i in range(k):
    # alternating logic

3. Not Considering Edge Cases

Failing to handle edge cases like k=1 or k=n-1 can lead to incorrect results.

Problematic Assumption:

# Assuming we always need both phases
def constructArray(self, n: int, k: int) -> List[int]:
    # ... Phase 1 logic ...
    # Might fail if k == n-1 (no elements left for Phase 2)

Robust Solution: The provided solution naturally handles these cases:

• When k=1: Only one alternation happens, then all remaining elements are added sequentially
• When k=n-1: The entire array is built using alternation (Phase 2 loop doesn't execute)

4. Misunderstanding the Problem Requirements

Some might think they need to maximize or minimize differences, or that the differences must be consecutive integers (1, 2, 3, ..., k).

Clarification: The problem only requires exactly k distinct difference values - they don't need to be any specific values or in any specific pattern. The alternating approach naturally creates differences of approximately n-1, n-2, n-3, etc., which satisfies the requirement.